Spivak 5.1 (iv) alternate solution

Question

This is in edition 3 of Spivak

The way I found a $\delta$ was much different than in the book and was wondering if it was correct (the value of my $\delta$ ended up being different as well):

Find a $\delta$ such that $|f(x) - l$ < $\epsilon$ $\forall$ x satisfying $0 < |x - a| < \delta$:

Proof:

$f(x) = \sqrt x ; a = 1; l = 1$

$|\sqrt x - 1|$ $\le$ $|x - 1|$ < $\delta$ = $\epsilon$

$\therefore$ $\delta = \epsilon$

Answer 1

Yes, you're right, as long as we make sure that $\delta\le 1$. The justification is that $$\left|\sqrt x-1\right| = \frac{|x-1|}{\sqrt x+1} \le |x-1|,$$ since $\sqrt x+1 \ge 1$.

By the way, there's never a unique "right answer" to such questions.